Compare int and Object in Java

Question

I have the following code:

Object obj = 3;
//obj.equals(3); // so is this true?

Does obj equal to 3?

Solution 1

What's at play here is autoboxing.

When you use a primitive literal when a reference is expected, the primitive is autoboxed to the wrapper type (in this case from int to Integer).

Your code is the equivalent of this:

Object obj = Integer.valueOf(3);
if ( obj.equals(Integer.valueOf(3)) ) {
    //...

I'll leave it to you to decide whether that's true or not.

Solution 2

This is also interesting:

Object t = 3;

t.equals( 3 );  // true
3 == t;         // true 

But

Object h = 128; 

h.equals( 128 ); // true 
128 == h;        // false

.equals will work, becase the value will be compared. == Will work, using the references, but only from -128 to 127, because the autoboxing mechanism, uses an internal pool to hold "most commonly used" references.

Strange enough: o == 3 will fail at compile time.

Solution 3

Yes.

Here is what's happening behind the scenes.

Object obj = Integer.valueOf(3);
obj.equals(Integer.valueOf(3));

So, of course they are equal.

Solution 4

The first statement will set obj to be a automatically boxed Integer (the same as Integer.valueOf(3))

Hence the second statement will return true.

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Comments

[Mark Peters]

What happened when you tried it?

[user]

obj.equals(3) was true, but I'd like to know if it was only "luck" or it really equals to 3

[Tony Ennis]

You're thinking along the right lines. Depending on luck sucks.

[medloh]

o? Where is that declared?